Choosing the Resistance in a Photoconductive Circuit

[Higgy]

Homework Statement

What should I choose for the resistor R_{F} and capacitor C_{F} in this circuit?

I have an unmounted silicon photodiode, with specs:

Reverse Voltage = 100 V
Short Circuit Current = 50 \mu A (typical)
Dark Current = 0.5 n A (typical), 1.0 n A (max)
Shunt Resistance = 250 M \Omega (min), 500 M \Omega (typical)
Junction Capacitance = 8 pF (typical)
Breakdown Voltage = 75 V (typical)

I'm trying to figure out how to properly use it. Laser light from an ~800nm diode laser will be incident on it, and I would like to view its output signal on an oscilloscope via a BNC coaxial cable.

I've learned that I can operate the photodiode in "photovoltaic mode", where there is zero bias and low dark current, but slow response time, or "photoconductive mode", where there is a negative bias and fast response time, but higher dark current.

Due to the circumstances of my experiment, I'm choosing to go with "photoconductive mode". Hence, the above circuit has a negative bias. I got it from AP Technologies website, where I also saw some useful formulas:

Homework Equations

V_{OUT} = I_{S} R_{F}

where I_{S} is the photodiode light signal current and R_{F} is the feedback resistance.

C_{F} = \frac{1}{2} \pi f R_{F}

where C_{F} is the feedback capacitance and f is the maximum operating frequency.

The Attempt at a Solution

I guess I'm just not sure what to use for values for calculating the resistance and capacitance needed. I've searched around online, and even though I've found some useful documents on this, I'm still left with two questions:

1. How large of a resistor do I need to not fry the diode?

2. What capacitor should I use? What is the "maximum operational frequency" referring to? Is it related to the RC time constant, or to the frequency of the incident light, etc?

My naive guess for the first one would be to use V_{OUT} \sim 10 V and I_{S} \sim 10 \mu A so that

R_{F} = \frac{10 V}{10 \mu A} = 1 M \Omega

That's big, though!

 

[Redbelly98]

R_F will depend on what current you expect for I_S, and how much signal you would like to have. I_S will depend on the amount of power from the diode beam hitting the photodiode's active area. You might take 1V as a reasonable signal level.

This circuit might work fine with no capacitor C_F. But if you use one, the question to answer is how fast must the output signal respond? Choose C_F so that the RC time constant is 1/3 of that time or less.

 

[Higgy]

R_F will depend on what current you expect for I_S, and how much signal you would like to have. I_S will depend on the amount of power from the diode beam hitting the photodiode's active area. You might take 1V as a reasonable signal level.

This circuit might work fine with no capacitor C_F. But if you use one, the question to answer is how fast must the output signal respond? Choose C_F so that the RC time constant is 1/3 of that time or less.

Thanks Redbelly98. Ignoring the capacitor, then, I'm still slightly confused as to what 'spec' value I should be looking at to help me choose a resistor that would not fry the diode.

Maybe my concern is unwarranted. In reverse-biasing the diode, we're not causing additional current flow through the diode, we're just setting up a potential difference, and the result is more output current from the photodiode? And it's current that would, in my words, "fry" the photodiode.

Since the capacitor-free circuit would just be an inverting amplifier, for which:

V_{out} = - \frac{R_{F}}{R_{D}} V_{in}

where R_{D} is the resistance of the diode, then with Ohm's law we could write:

I_{S} R_{D} = -\frac{R_{D}}{R_{F}} V_{out} \rightarrow R_{F} = -\frac{V_{out}}{I_{S}}

and with I_{S} = 50 \mu A and V_{out} = 1 V we have

R_{F} = 50 M \Omega

Is that roughly the right way to think about it?

 

[Redbelly98]

1V / 50 μA is 20kΩ, not 50MΩ. 20kΩ sounds reasonable.

Forget about "resistance of the diode", instead think in terms of diode current. Actually, this is essentially what you are doing since your R_D terms canceled out of the equation.

The photodiode is essentially a current source here. At 800 nm, it can generate 0.5-1.0 Amps per Watt of optical power. To generate 50 μA, it would be receiving 50-100 μW from the laser. This should not hurt the photodiode.

I'll also suggest building a preliminary circuit and experimenting to find the right resistor value. No op-amp, just a loop circuit containing the photodiode, biasing voltage, and resistor. A simple 1.5V battery (or two of them) can be used for bias. Then try different resistor values until you get a volt or so of signal across the resistor.

p.s. if the resistor voltage is close to or larger than the bias voltage, then you are saturating the photodiode. Increase the bias voltage or reduce the resistor so that the resistor voltage is less than the bias.

p.p.s. For really fast-changing signals, say in the RF frequency range (several ns or faster), it is common practice to use a 50Ω resistor with the photodiode, and then use amplifiers to amplify the voltage across the 50Ω resistor.

 

[Higgy]

1V / 50 μA is 20kΩ, not 50MΩ. 20kΩ sounds reasonable.

Oops. Right, that makes more sense.

Forget about "resistance of the diode", instead think in terms of diode current. Actually, this is essentially what you are doing since your R_D terms canceled out of the equation.

The photodiode is essentially a current source here. At 800 nm, it can generate 0.5-1.0 Amps per Watt of optical power. To generate 50 μA, it would be receiving 50-100 μW from the laser. This should not hurt the photodiode.

I'll also suggest building a preliminary circuit and experimenting to find the right resistor value. No op-amp, just a loop circuit containing the photodiode, biasing voltage, and resistor. A simple 1.5V battery (or two of them) can be used for bias. Then try different resistor values until you get a volt or so of signal across the resistor.

p.s. if the resistor voltage is close to or larger than the bias voltage, then you are saturating the photodiode. Increase the bias voltage or reduce the resistor so that the resistor voltage is less than the bias.

p.p.s. For really fast-changing signals, say in the RF frequency range (several ns or faster), it is common practice to use a 50Ω resistor with the photodiode, and then use amplifiers to amplify the voltage across the 50Ω resistor.

Thanks! This really clears things up. I'll try your suggestion of circuit-sans-op-amp, and go from there.